{% endfor %}
diff --git a/source/infra/css/main.css b/source/infra/css/main.css
index cdf71e4..242c08d 100644
--- a/source/infra/css/main.css
+++ b/source/infra/css/main.css
@@ -37,10 +37,14 @@ body {
font-family: sans-serif;
}
-h1, h2, h3 {text-align: center;}
-h1 {font-size: 2.0rem;}
-h2 {font-size: 1.6rem;}
-h3 {font-size: 1.2rem;}
+h1, h2, h3, h4 {text-align: center;}
+h1 {font-size: 2.1rem;}
+h2 {font-size: 1.8rem;}
+h3 {font-size: 1.5rem;}
+h4 {
+ font-size: 1.2rem;
+ font-style: italic;
+}
a {
background-color: transparent;
text-decoration: none;
diff --git a/source/infra/css/syntax.css b/source/infra/css/syntax.css
index 50191d2..08e24b4 100644
--- a/source/infra/css/syntax.css
+++ b/source/infra/css/syntax.css
@@ -1,4 +1,5 @@
.highlighter-rouge {
+ padding: 0 2px;
border: 1px solid;
border-color: #999988;
background-color: #f8f8f8;
diff --git a/source/know/concept/boltzmann-relation/index.md b/source/know/concept/boltzmann-relation/index.md
index d5409d2..b528adf 100644
--- a/source/know/concept/boltzmann-relation/index.md
+++ b/source/know/concept/boltzmann-relation/index.md
@@ -71,15 +71,16 @@ $$\begin{aligned}
}
\end{aligned}$$
-However, due to their larger mass,
-ions are much slower to respond to fluctuations in the above equilibrium.
+But due to their large mass,
+ions respond much slower to fluctuations in the above equilibrium.
Consequently, after a perturbation,
-the ions spend much more time in a transient non-equilibrium state
+the ions spend more time in a transient non-equilibrium state
than the electrons, so this formula for $$n_i$$ is only valid
if the perturbation is sufficiently slow,
-allowing the ions to keep up.
+such that the ions can keep up.
Usually, electrons do not suffer the same issue,
-thanks to their small mass and fast response.
+thanks to their small mass and hence fast response.
+
## References
diff --git a/source/know/concept/coulomb-logarithm/index.md b/source/know/concept/coulomb-logarithm/index.md
index b843eb3..b3be5ac 100644
--- a/source/know/concept/coulomb-logarithm/index.md
+++ b/source/know/concept/coulomb-logarithm/index.md
@@ -147,12 +147,12 @@ We thus find:
$$\begin{aligned}
\boxed{
\sigma_\mathrm{small}
- = 8 \ln(\Lambda) \sigma_\mathrm{large}
- = \frac{q_1^2 q_2^2 \ln\!(\Lambda)}{2 \pi \varepsilon_0^2 |\vb{v}|^4 \mu^2}
+ = 8 \sigma_\mathrm{large} \ln(\Lambda)
+ = \frac{q_1^2 q_2^2 \ln(\Lambda)}{2 \pi \varepsilon_0^2 |\vb{v}|^4 \mu^2}
}
\end{aligned}$$
-Here, $$\ln\!(\Lambda)$$ is known as the **Coulomb logarithm**,
+Here, $$\ln(\Lambda)$$ is known as the **Coulomb logarithm**,
with the **plasma parameter** $$\Lambda$$ defined below,
equal to $$9/2$$ times the number of particles
in a sphere with radius $$\lambda_D$$:
@@ -168,7 +168,7 @@ $$\begin{aligned}
The above relation between $$\sigma_\mathrm{small}$$ and $$\sigma_\mathrm{large}$$
gives us an estimate of how much more often
small deflections occur, compared to large ones.
-In a typical plasma, $$\ln\!(\Lambda)$$ is between 6 and 25,
+In a typical plasma, $$\ln(\Lambda)$$ is between 6 and 25,
such that $$\sigma_\mathrm{small}$$ is 2-3 orders of magnitude larger than $$\sigma_\mathrm{large}$$.
Note that $$t$$ is now fixed as the period
@@ -179,7 +179,7 @@ for significant energy transfer between partices:
$$\begin{aligned}
\frac{1}{t}
= n |\vb{v}| \sigma_\mathrm{small}
- = \frac{q_1^2 q_2^2 \ln\!(\Lambda) \: n}{2 \pi \varepsilon_0^2 \mu^2 |\vb{v}|^3}
+ = \frac{q_1^2 q_2^2 \ln(\Lambda) \: n}{2 \pi \varepsilon_0^2 \mu^2 |\vb{v}|^3}
\sim \frac{n}{T^{3/2}}
\end{aligned}$$
diff --git a/source/know/concept/fabry-perot-cavity/index.md b/source/know/concept/fabry-perot-cavity/index.md
index 6eefc6e..c013f1d 100644
--- a/source/know/concept/fabry-perot-cavity/index.md
+++ b/source/know/concept/fabry-perot-cavity/index.md
@@ -50,6 +50,7 @@ $$\begin{aligned}
A_4 e^{i k_m n_R \ell/2}
&= A_2 e^{- i k_m n_C \ell/2} + A_3 e^{i k_m n_C \ell/2}
\end{aligned}$$
+
$$\begin{aligned}
- i k_m n_L A_1 e^{i k_m n_L \ell/2}
&= - i k_m n_C A_2 e^{i k_m n_C \ell/2} + i k_m n_C A_3 e^{- i k_m n_C \ell/2}
diff --git a/source/know/concept/ion-sound-wave/index.md b/source/know/concept/ion-sound-wave/index.md
index cb86c04..8749f1a 100644
--- a/source/know/concept/ion-sound-wave/index.md
+++ b/source/know/concept/ion-sound-wave/index.md
@@ -95,16 +95,16 @@ we make the following plane-wave ansatz:
$$\begin{aligned}
n_{i1}(\vb{r}, t)
- &= n_{i1} \exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+ &= n_{i1} \exp(i \vb{k} \cdot \vb{r} - i \omega t)
\\
n_{e1}(\vb{r}, t)
- &= n_{e1} \exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+ &= n_{e1} \exp(i \vb{k} \cdot \vb{r} - i \omega t)
\\
\vb{u}_{i1}(\vb{r}, t)
- &= \vb{u}_{i1} \exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+ &= \vb{u}_{i1} \exp(i \vb{k} \cdot \vb{r} - i \omega t)
\\
\phi_1(\vb{r}, t)
- &= \phi_1 \,\,\exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+ &= \phi_1 \,\,\exp(i \vb{k} \cdot \vb{r} - i \omega t)
\end{aligned}$$
Which we then insert into the momentum equations for the ions and electrons:
diff --git a/source/know/concept/kramers-kronig-relations/index.md b/source/know/concept/kramers-kronig-relations/index.md
index 3880113..711023e 100644
--- a/source/know/concept/kramers-kronig-relations/index.md
+++ b/source/know/concept/kramers-kronig-relations/index.md
@@ -58,7 +58,8 @@ $$\begin{aligned}
\pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}}
\end{aligned}$$
-From the definition of the Fourier transform we know that $$2 \pi A B / |s| = 1$$:
+From the definition of the Fourier transform we know that
+$$2 \pi A B / |s| = 1$$:
$$\begin{aligned}
\tilde{\chi}(\omega)
diff --git a/source/know/concept/langmuir-waves/index.md b/source/know/concept/langmuir-waves/index.md
index 7dc5dbf..be47567 100644
--- a/source/know/concept/langmuir-waves/index.md
+++ b/source/know/concept/langmuir-waves/index.md
@@ -130,8 +130,8 @@ where we neglect $$(\vb{u}_{e1} \cdot \nabla) \vb{u}_{e1}$$
because $$\vb{u}_{e1}$$ is so small by assumption:
$$\begin{gathered}
- m_e (n_{e0} \!+\! n_{e1}) \Big( \pdv{(\vb{u}_{e0} \!+\! \vb{u}_{e1})}{t}
- + \big( (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \cdot \nabla \big) (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \Big)
+ m_e (n_{e0} \!+\! n_{e1}) \bigg( \pdv{(\vb{u}_{e0} \!+\! \vb{u}_{e1})}{t}
+ + \big( (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \cdot \nabla \big) (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \bigg)
= q_e \big( n_{e0} \!+\! n_{e1} \big) \big( \vb{E}_0 \!+\! \vb{E}_1 \big)
\\
\implies \qquad
diff --git a/source/know/concept/laser-rate-equations/index.md b/source/know/concept/laser-rate-equations/index.md
index a84d274..1f42f73 100644
--- a/source/know/concept/laser-rate-equations/index.md
+++ b/source/know/concept/laser-rate-equations/index.md
@@ -187,7 +187,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-To rewrite this, we replace $$|\vb{E}|^2$$ with the photon number $$N_p$$ as follows,
+To rewrite this, we replace $$|\vb{E}|^2$$
+with the photon number $$N_p$$ as follows,
with $$U = \varepsilon_0 n^2 |\vb{E}|^2 / 2$$ being the energy density of the light:
$$\begin{aligned}
diff --git a/source/know/concept/lehmann-representation/index.md b/source/know/concept/lehmann-representation/index.md
index 74bd457..0dec718 100644
--- a/source/know/concept/lehmann-representation/index.md
+++ b/source/know/concept/lehmann-representation/index.md
@@ -40,7 +40,7 @@ $$\begin{aligned}
\Matrixel{n}{e^{i \hat{H} (t - t') / \hbar} \hat{c}_\nu e^{- i \hat{H} (t - t') / \hbar} \hat{c}_{\nu'}^\dagger}{n}
\end{aligned}$$
-Where we used that the trace $$\Tr\!(x) = \sum_{n} \matrixel{n}{x}{n}$$
+Where we used that the trace $$\Tr(x) = \sum_{n} \matrixel{n}{x}{n}$$
is invariant under cyclic permutations of $$x$$.
The $$\Ket{n}$$ form a basis of eigenstates of $$\hat{H}$$,
so we insert an identity operator $$\sum_{n'} \Ket{n'} \Bra{n'}$$:
diff --git a/source/know/concept/martingale/index.md b/source/know/concept/martingale/index.md
index 9d3c6b4..53a346a 100644
--- a/source/know/concept/martingale/index.md
+++ b/source/know/concept/martingale/index.md
@@ -45,12 +45,14 @@ Modifying property (3) leads to two common generalizations.
The stochastic process $$M_t$$ above is a **submartingale**
if the current value is a lower bound for the expectation:
-3. For $$0 \le s \le t$$, the conditional expectation $$\mathbf{E}(M_t | \mathcal{F}_s) \ge M_s$$.
+3. For $$0 \le s \le t$$, the conditional expectation
+ $$\mathbf{E}(M_t | \mathcal{F}_s) \ge M_s$$.
Analogouly, $$M_t$$ is a **supermartingale**
if the current value is an upper bound instead:
-3. For $$0 \le s \le t$$, the conditional expectation $$\mathbf{E}(M_t | \mathcal{F}_s) \le M_s$$.
+3. For $$0 \le s \le t$$, the conditional expectation
+ $$\mathbf{E}(M_t | \mathcal{F}_s) \le M_s$$.
Clearly, submartingales and supermartingales are *biased* random walks,
since they will tend to increase and decrease with time, respectively.
diff --git a/source/know/concept/maxwell-boltzmann-distribution/index.md b/source/know/concept/maxwell-boltzmann-distribution/index.md
index ebb2460..318e659 100644
--- a/source/know/concept/maxwell-boltzmann-distribution/index.md
+++ b/source/know/concept/maxwell-boltzmann-distribution/index.md
@@ -81,7 +81,7 @@ after normalization:
$$\begin{aligned}
f(p_x, p_y, p_z)
- = \Big( \frac{1}{2 \pi m k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{(p_x^2 + p_y^2 + p_z^2)}{2 m k_B T} \Big)
+ = \Big( \frac{1}{2 \pi m k_B T} \Big)^{3/2} \exp\!\bigg( \!-\!\frac{(p_x^2 + p_y^2 + p_z^2)}{2 m k_B T} \bigg)
\end{aligned}$$
We now rewrite this using the velocities $$v_x = p_x / m$$,
@@ -90,7 +90,7 @@ and update the normalization, giving:
$$\begin{aligned}
\boxed{
f(v_x, v_y, v_z)
- = \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{m (v_x^2 + v_y^2 + v_z^2)}{2 k_B T} \Big)
+ = \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} \exp\!\bigg( \!-\!\frac{m (v_x^2 + v_y^2 + v_z^2)}{2 k_B T} \bigg)
}
\end{aligned}$$
diff --git a/source/know/concept/random-phase-approximation/index.md b/source/know/concept/random-phase-approximation/index.md
index 698e1e7..0d0b428 100644
--- a/source/know/concept/random-phase-approximation/index.md
+++ b/source/know/concept/random-phase-approximation/index.md
@@ -24,7 +24,7 @@ and $$k_F^2$$ from the energy $$1 / \beta$$:
$$\begin{aligned}
\frac{1}{(2 \pi)^3} \int_{-\infty}^\infty \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty \cdots \:\dd{\vb{k}}
- \:\:\sim\:\:
+ \quad\sim\quad
k_F^5
\end{aligned}$$
diff --git a/source/know/concept/self-steepening/index.md b/source/know/concept/self-steepening/index.md
index f934ab7..9666167 100644
--- a/source/know/concept/self-steepening/index.md
+++ b/source/know/concept/self-steepening/index.md
@@ -19,7 +19,7 @@ nonlinear Schrödinger equation:
$$\begin{aligned}
0
- = i\pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma \Big(1 + \frac{i}{\omega_0} \pdv{}{t} \Big) \big(|A|^2 A\big)
+ = i\pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma \Big(1 + \frac{i}{\omega_0} \pdv{}{t} \Big) |A|^2 A
\end{aligned}$$
Where $$\omega_0$$ is the angular frequency of the pump.
diff --git a/source/know/concept/stokes-law/index.md b/source/know/concept/stokes-law/index.md
index 3a02a83..40212ef 100644
--- a/source/know/concept/stokes-law/index.md
+++ b/source/know/concept/stokes-law/index.md
@@ -50,10 +50,10 @@ $$\begin{gathered}
\\
v_r
= U f(r) \cos\theta
- \qquad
+ \qquad \quad
v_\theta
= - U g(r) \sin\theta
- \qquad
+ \qquad \quad
v_\phi
= 0
\end{gathered}$$
@@ -76,7 +76,7 @@ $$\begin{aligned}
\\
&= U \dv{f}{r} \cos\theta - \frac{U g}{r} \cos\theta + \frac{2 U f}{r} \cos\theta - \frac{U g}{r} \cos\theta
\\
- &= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big)
+ &= U \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big) \cos\theta
\end{aligned}$$
The parenthesized expression must be zero for all $$r$$,
@@ -103,16 +103,16 @@ which is as follows for our ansatz $$p(r, \theta)$$:
$$\begin{aligned}
0
= \nabla^2 p
- &= \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin\theta \pdv{p}{\theta} \Big)
+ &= \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \pdv{p}{\theta} \sin\theta \Big)
\\
0
&= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big)
- \frac{\eta U q}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin^2\theta \Big)
\\
- &= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big)
+ &= \frac{\eta U}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big) \cos\theta
- \frac{2 \eta U q}{r^2 \sin\theta} \sin\theta \cos\theta
\\
- &= \eta U \cos\theta \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big)
+ &= \eta U \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big) \cos\theta
\end{aligned}$$
Again, the parenthesized expression must be zero for all $$r$$,
@@ -129,7 +129,7 @@ The pressure is therefore:
$$\begin{aligned}
p
- = \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big)
+ = \eta U \Big( \frac{C_3}{r^2} + C_4 r \Big) \cos\theta
\end{aligned}$$
Consequently, its gradient $$\nabla p$$ in spherical coordinates is as follows:
@@ -137,7 +137,7 @@ Consequently, its gradient $$\nabla p$$ in spherical coordinates is as follows:
$$\begin{aligned}
\nabla p
= \vu{e}_r \pdv{p}{r} + \vu{e}_\theta \frac{1}{r} \pdv{p}{\theta}
- = \vu{e}_r \Big( \eta U \cos\theta \dv{q}{r} \Big) - \vu{e}_\theta \Big( \eta U \sin\theta \frac{q}{r} \Big)
+ = \vu{e}_r \Big( \eta U \dv{q}{r} \cos\theta \Big) - \vu{e}_\theta \Big( \eta U \frac{q}{r} \sin\theta \Big)
\end{aligned}$$
According to the Stokes equation, this equals $$\eta \nabla^2 \va{v}$$.
@@ -148,24 +148,24 @@ $$\begin{aligned}
&= \pdvn{2}{v_r}{r} + \frac{1}{r^2} \pdvn{2}{v_r}{\theta} + \frac{2}{r} \pdv{v_r}{r}
+ \frac{\cot\theta}{r^2} \pdv{v_r}{\theta} - \frac{2}{r^2} \pdv{v_\theta}{\theta} - \frac{2}{r^2} v_r - \frac{2 \cot\theta}{r^2} v_\theta
\\
- &= U \cos\theta \Big( \dvn{2}{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f
- + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big)
+ &= U \Big( \dvn{2}{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f
+ + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big) \cos\theta
\\
- &= U \cos\theta \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big)
+ &= U \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big) \cos\theta
\end{aligned}$$
Substituting $$g$$ for the expression we found from incompressibility lets us simplify this:
$$\begin{aligned}
\eta (\nabla^2 \va{v})_r
- &= \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
+ &= \eta U \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) \cos\theta
\end{aligned}$$
The Stokes equation says that this must be equal to the $$r$$-component of $$\nabla p$$:
$$\begin{aligned}
- \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
- = \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big)
+ \eta U \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big) \cos\theta
+ = \eta U \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big) \cos\theta
\end{aligned}$$
Where we have inserted $$\idv{q}{r}$$.
@@ -213,14 +213,15 @@ $$\begin{gathered}
\\
\boxed{
v_r
- = U \cos\theta \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big)
+ = U \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big) \cos\theta
\qquad
v_\theta
- = - U \sin\theta \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big)
+ = - U \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big) \sin\theta
}
\end{gathered}$$
+
## Drag force
From the definition of [viscosity](/know/concept/viscosity/),
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